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Under what conditions are the products of inertia (off-diagonal elements of inertia tensor) non-zero? It seems that for many objects, constructing the moment of inertia tensor results in something like the following:

\begin{matrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0&0 & I_{zz} \end{matrix}

From a glance, it seems like this is due to some agreement between the distribution of the objects mass, and the way the co-ordinate system has been aligned. For example, we'd align our axes parallel to the sides of a rectangular prism. Moving the origin by the parallel axis theorem often introduces non-zero product of inertia terms, as well.

What specifically causes this difference? Is there always a set of axes for any object, rotating around any point where they are zero? Is this a property of rotating around some point other than the center of gravity?

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    $\begingroup$ There is a theorem that says you can find a coordinate system for which the inertial tensor is diagonal for any mass distribution. Look in any any graduate mechanics text and most upper-division ones. So ... under all conditions other than the ones needed to achieve diagonalization? $\endgroup$ – dmckee Jun 3 '14 at 3:41
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    $\begingroup$ Related: physics.stackexchange.com/q/66350 $\endgroup$ – dmckee Jun 3 '14 at 3:44

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